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Introduction to Symbolic LogicAnswer Key

Dr. Daniel R. Kern

Chapter 1

A. State whether the following sentences are statements or not. If not, say what kind of sentence it is.

1. Statement.2. Non-statement, question.3. Statement (even though we don’t know whether it is true or false, it IS either true or false.)4. Statement (even though we don’t know whether it is true or false, it IS either true or false.)5. Statement.6. Non-statement, command.7. Statement.8. Although this is exclamatory in form, the team either won the game or it didn’t, so this is a

statement.9. Non-statement, exclamation.10. Statement.

B. State whether the following statements are fact or value statements. Then, using the criterion in the chapter, determine whether each is an objective or subjective statement.

1. Value statement – evaluative. Subjective – people don’t have real arguments about the color of eye shadow.

2. Value statement – all “should” statements are evaluative. Objective – people in the U.S. did and still do argue about whether this statement is true or false.

3. Fact statement. Objective – people could argue about this.4. Value statement. Whether it is subjective or objective depends on whether the word “best” is

defined according to particular criteria (most games won, most seasons won, etc), in which case it would be objective, or whether it just a statement of emotional attachment to the Lakers, in which case it would be subjective.

5. Value statement – moral statement. Objective, because people argue about the truth of this statement.

6. Fact statement. Objective.7. Value statement – moral statement. Objective, because people argue about the truth of this

statement.8. Fact statement (even though we don’t know whether it is true or false). Objective.9. Fact statement (even though we don’t know whether it is true or false). Objective.10. This is the best possibility for a factual subjective statement. It is a statement of a preference, so

could be taken as a factual statement (there is a fact about my particular tastes and this statement says something about the world). But the statement would be true if I said it, but would be false if someone who didn’t like butterscotch ice cream said it.

C. Evaluate the following arguments:

1. Argument by Authority, inductive. Weak because there is no evidence that this person is an authority on elections. Question of cogency doesn’t apply.

2. Argument by definition, deductive. Invalid. If the premise is true, the conclusion IS false. Question of soundness doesn’t apply.


Dr. Daniel R. Kern

3. Statistical argument, inductive. Weak. Even if these friends said that it wasn’t dangerous to drive after drinking, they are a small and unrepresentative group, so the conclusion is still very improbable.

4. Causal argument, inductive. Strong. IF the premises were true, the conclusion would be probable. Uncogent. It is false that a black cat crossing one’s path affects one’s luck.

5. Categorical argument, deductive. Valid. IF the premises were true, the conclusion would necessarily be true. Unsound. It is false that all mammals have 15 hearts.

6. Analogy, inductive. Strong. Although it is possible that Bill wouldn’t be able to do 100 pushups, just because he, like Bob was just discharged from the Navy Seals, strength is one of the characteristics of Navy Seals, so it is probable that he could do the pushups.

7. Statistical argument, inductive. Relatively strong. There is in fact a 67% chance that any particular person would favor abolishing the death penalty, if the premise is true.

8. Mathematical argument, deductive. Valid. If the premises are true, the conclusion would necessarily be true as well. There is nothing obviously false about the premises, so we would take them to be true, so it would be sound as well.

9. Causal argument, inductive. Strong. There is a strong causal connection between leaving your headlights on and the car battery being dead. There is nothing obviously false about the premises, so we would take them to be true, so it would be cogent as well.

10. Categorical argument, deductive. Valid. If the premises are true, the conclusion would necessarily be true as well. The first premise is true and there is nothing obviously false about the second premise, so we would take it to be true, so it would be sound as well

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Dr. Daniel R. Kern

Chapter 2

Translation Exercises

M = Mary sings, B = Bob sings, C = Mary dances, D = Bob dances 1. (M & D)2. (M & B) & (C & D) (the order and grouping make no difference on this one.3. M C 4. ~(M D)5. M C6. (B M) (C & D)7. (B D) (M & ~C)8. (B M) & (D M)9. (M B) & (D C)10. (~M ~B) & (D ~C)11. (~C ~D) & (~B ~M)12. M ~B13. (C & D) (M & B)14. M & (C ~(B D))15. (M & C) ~(B D)16. ((M & D) ~B) & ~D17. ((M B) C) & (D (M & ~B))18. (M D) (B ~C)19. ~[(M & B) (C & D)] & [~(M B) ~(C D)]20. ~(M D) ~[(M & B) (~C & ~D)]

Partial Truth TablesTruth Values:

M = T, B = T, C = T, D = F

1. (M & D)T F F

2. (M & B) & (C & D) T T T F T F F

3. M C T T T

4. ~(M D)T T F F


Dr. Daniel R. Kern

5. (B M) CT T T T T

6. (B M) (C & D)T T T F T T F

7. (B D) (M & ~C)T T T F T F FT

8. (B M) & (D M)T T T T F T T

9. (M B) & (D C)T T T T F T T

10. (~M ~B) & (D ~C)FT F FT F F T FT

11. (~C ~D) & (~B ~M)FT T TF T FT T FT

12. M ~BT F FT

13. (C & D) (M & B)T F F F T T T

14. M & (C ~(B D))T F T F F T T F

15. (M & C) ~(B D)T T T F F T T F

16. ((M & D) ~B) & ~DT F F T FT T TF

17. ((M B) C) & (D (M & ~B))T T T T T F F T T F FT

18. (M D) (B ~C)T F F T T F FT

19. ~[(M & B) (C & D)] & [~(M B) ~(C D)]T T T T F T F F T F T T T T F T T F

20. ~(M D) ~[(M & B) (~C & ~D)]T T F F F F T T T T FT F TF

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Dr. Daniel R. Kern

Full Truth Tables for single statements

1. A B (A & B) & (A & ~B) T T T T T F T F FTT F T F F F T T TFF T F F F F F F FTF F F F F F F F TF All false - contradiction

2. A B (A B) (A & ~B) T T T T T F T F FTT F T F F T T T TFF T F T T F F F FTF F F T F F F F TF Both Ts and Fs - Contingent

3. A B (A B) (~A ~B) T T T T T T FT F FTT F T T F T FT T TFF T F T T T TF T FTF F F F F T TF T TF Tautology – All true.

4. A B C ((A B) C) & ~C T T T T T T T T F FTT T F T T T F F F TFT F T T T T T T F FTT F F T F F F F F TFF T T F T T T T F FTF T F F T T T T T TF There are now both Ts F F T and Fs on the table, so F F F it can be declared

contingent.

5. M D B (M & D) (~B ~D) T T T T T T F FT F FT T T F T T T F TF F FTT F T T F F F FT T TFT F F T F F F TF T TFF T T F F T T FT F FT There are now both TsF T F and Fs on the table, soF F T it can be declaredF F F contingent.

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Dr. Daniel R. Kern

Full Truth Tables for pairs of statements

1. A B A B ~A B T T T T T FT T TT F T F F FT F FF T F T T TF T TF F F T F TF T F Equivalent

2. A B ~(A B) ~A & ~B T T F T T T FT F FTT F F T T F FT F TFF T F F T T TF F FTF F T F F F TF T TF Equivalent

3. A B ~(A & B) ~A ~B T T F T T T FT F FTT F T T F F FT T TFF T T F F T TF T FTF F T F F F TF T TF Equivalent

4. A B C A (B C) (A B) C T T T T T T T T T T T T TT T F T F T F F T T T F FT F T T T F T T T F T T TT F F T T F T F T F F T FF T T F T T T T F T T T TF T F F T T F F F T T F F Neither, notF F T equivalentF F F

5. A B C A (B C) (A & B) C T T T T T T T T T T T T TT T F T F T F F T T T F FT F T T T F T T T T T T TT F F T T F T F T F F T FF T T F T T T T F F T T TF T F F T T F F F F T T FF F T F T F T T F F F T FF F F F T F T F F F F T F Equivalent

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Dr. Daniel R. Kern

Truth tables for consistency

Use a full truth table to test each of the following sets of statements for consistency. Be able to explain what makes the statements consistent or inconsistent.

NOTE: For these problems, I have placed the Ts and Fs directly under the statement letters

1. A & ~B A B A ~B T F FT T T T T F FTT T TF T T F T T TF ConsistentF F FT F T T F T FTF F TF F F F F T TF

2. A (A B) (A & ~B) (A & B) A & B T T T T T T F FT F T T T T T TT T T T F T T TF F T F T T F FF F F T T F F FT T F F T F T TF T F F F F F TF T F F F F F F

Inconsistent – There are NO rows (horizontal) on which the statements are ALL true.

3. A (A B) A & (A & ~B) A B T T T T T T F T F FT T T TT T T T F T T T T TF T F FF T F T T F F F F FT F T TF F F F F F F F F TF F T F

Inconsistent – There are NO rows (horizontal) on which the statements are ALL true.

4. A (B C) (A & B) ~C B & ~C T T T T T T T T F FT T F FTT T T F F T T T T TF T T TF ConsistentT T F T T T F F T FT F F FTT T F T F T F F T TF F F TFF T T T T F F T T FT T F FTF T T T F F F T T TF T T TFF T F T T F F F T FT F F FTF F F F F F F F T TF F F TF

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Dr. Daniel R. Kern

5. B (C & D) ~B & (C & D) B (C D) T T T T T FT F T T T T T T T TT F T F F FT F T F F T T T T FT F F F T FT F F F T T T F T TT F F F F FT F F F F T T F T FF T T T T TF T T T T F T T F T ConsistentF T T T F TF F T F F F T T T FF T F T T TF F F F T F T F T TF F F F F TF F F F F F F F F F

Truth tables for Validity

NOTE: For these problems, I have placed the Ts and Fs directly under the statement letters

6. A & ~B A B A ~B T F FT T T T T F FTT T TF T T F T T TFF F FT F T T F T FTF F TF F F F F T TF

Valid – there are NO rows on which the two premises are true and the conclusion is false.

7. A (A B) (A & ~B) (A & B) A & B T T T T T T F FT F T T T T T TT T T T F T T TF F T F T T F FF F F T T F F FT T F F T F T TF T F F F F F TF T F F F F F F

Invalid – There is a row (the 4th row) on which the premises are true and the conclusion is false.

8. A (A B) A & (A & ~B) A B T T T T T T F T F FT T T TT T T T F T T T T TF T F FF T F T T F F F F FT F T TF F F F F F F F F TF F T F

Invalid – There is a row (the 2nd row) on which the premises are true and the conclusion is false.

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Dr. Daniel R. Kern

9. A (B C) (A & B) ~C B & ~C T T T T T T T T F FT T F FTT T T F F T T T T TF T T TFT T F T T T F F T FT F F FTT T F T F T F F T TF F F TFF T T T T F F T T FT T F FTF T T T F F F T T TF T T TFF T F T T F F F T FT F F FTF F F F F F F F T TF F F TF

Invalid – There is a row (the 3rd row, among others) on which the premises are true and the conclusion is false.

10. B (C & D) ~B & (C & D) B (C D) T T T T T FT F T T T T T T T TT F T F F FT F T F F T T T T FT F F F T FT F F F T T T F T TT F F F F FT F F F F T T F T FF T T T T TF T T T T F T T F TF T T T F TF F T F F F T T T FF T F T T TF F F F T F T F T TF F F F F TF F F F F F F F F F

Valid – there are NO rows on which the premises are true and the conclusion is false.

Shortened truth tables for consistency

1. A (B C) (A & B) ~C B & ~C T T T T T FT T F FTT T T T F T T T T TF T T TFT F T T F FT F F FTT F F T F TF F F TFF T T F T FT T F FTF T F F T TF T T TFF F T F F FT F F FTF F F F F TF F F TF

Consistent. The third statement is false in all the rows but the 2nd and 6th. The first and second statements are also true on the second row, so there is a row on which all the statements are true.

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Dr. Daniel R. Kern

2. B (C & D) ~B & (C D) B (C D) T T T FT F T T T T TT T F FT F T F T T FT F T FT F F T T F TT F F FT F F F T F FF F T T T TF T T T T F T TF T F TF F T F F F T FF T F F T TF T F T T F T F T TF F F TF T F T F F F F

Consistent. The second statement is false in all the rows but the 5th, 7th, and 8th. The first statement is false on the 5th row, but all three statements are true on the 7th row, so there is a row on which all the statements are true.

3. A & (B & ~C) B (~A ~C) (C A) & ~B T F T F FT T FT FT T T FTT T T T TF T T FT TF F F T F FTT F F F FT F FT FT T T TFT F F F TF F FT TF F T TFF F T FT T TF FT T F FTF F T TF T TF TF T F FTF F F FT F TF FT F F TFF F F TF F TF TF F F TF

Inconsistent. The first statement is false on all rows except the 2nd row. On that row the third statement is false. So there are NO rows on which all of the statements are true.

4. (B & A) (A ~C) (A B) C (B & C) (A & B) T T T F T F FT T T T T T T TT T T T T T TF T T T F F T F T TF F T T T F FT T T F T T F F T T T F FF F T T T F TF T F F F F T FT F F F F T FT F T T T T F TT F F F F T TF F T F T F F TF F F F F T FT F F T F T F FF F F F F T TF F F F F F F F

Consistent – all three statements are true on the third line.

5. (A B) (B ~C) ((A & B) ~C) B (A & B) & ~C T T T FT T T FT T T T F FTT T T T T T TF T T T T TF T T T T T TFT F F FT T F FT F T F F F FTT F F TF T F TF F T F F F TFF T T FT F T FT T F F T F FTF T T TF F T TF T F F T F TFF F F FT F F FT F F F F F FTF F F TF F F TF F F F F F TF

Consistent – all three statements are true on the second line.

10.


Dr. Daniel R. Kern

Shortened truth tables for validity

1. A (B C) (A & B) ~C B & ~C T T T T T T F FT T F FTT T F T T TF T T TFT T F T T F F T FT F F FTT F F T F TF F F TFF T T F T FT T F FTF T F F T TF T T TFF F T F F FT F F FTF F F F F TF F F TF

Invalid – on the 3rd row, the premises are true while the conclusion is false.

2. B (C & D) ~B & (C D) B (C D) T T T FT T T T T T FTT T F FT T F T T T TFT F T FT F T T T F FTT F F FT F F T T F TFF T T TF T T F T T T FTF T F TF T F F T T T TFF F T TF F F T F F F F FTF F F TF F F F T F T TF

Valid – on the only row on which the conclusion is false, row 7, the second premise is false. So there are NO rows on which the premises are true while the conclusion is false.

3. A & (B & ~C) B (~A ~C) (C A) & ~B T F T F FT T FT FT T T F FTT T T TF T T FT T TF F T F FTT F FT F FT FT T T T T TFT F TF F FT TF F F T F TFF T FT T TF FT T F F FTF T TF T TF TF F F F FTF F FT F TF FT T F F F TFF F TF F TF TF F T F T TF

Invalid – on the second row, both premises are true while the conclusion is false.

4. (B & A) (A ~C) (A B) C (B & C) (A & B) T T T FT T T T T T T T T T TT T T TF T T F T F F T T T TF T T FT T F T F F T T T FF T T TF T F F F F F T T FT F F F F T FT F T T T T T T T F F F TT F F TF F T F T F F T F F TF F F FT F F T F F T T F F FF F F TF F F F F F F T F F F

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Dr. Daniel R. Kern

Valid – the only row on which the conclusion is false is the 5th row. On that row the first premise is false – so there are NO rows on which the premises are true while the conclusion is false.

5. (A B) (B ~C) ((A & B) ~C) B (A & B) & ~C T T T FT T T T F FT F T T T F FTT T T TF T T TF T T T T T TFT F F FT T F F T FT F F T F F FTT F F TF T F F T TF F F T F F F TFF T T F T F FT F F T T FT T T F T F FTF T T T T T TF F F T T TF T T F F T F TFF F F FT F F FT F F F F F FTF F F TF F F TF F F F F F TF

Invalid – on the 6th row the premises are all true while the conclusion is false.

The Indirect Method for testing for consistency

1. A & (B & ~C) B (~A ~C) (C A) & ~BT T T T TF T T FT T TF F F T T FT

Since it was possible to make all of the statements true at the same time, they are consistent.

2. (A B) (B ~C) ((A & B) C) B (A & B) & ~C T T T T T T TF T T T F F T T T T T T TF

There is only one way for the third statement to be true, if A and B are true, and C is false (~C is true). However, this truth value assignment makes the second statement false. It is impossible to make all of the statements true at the same time, so they are inconsistent.

3. (B & A) (A C) ~[(A & B) C] (B & C) (A & B)T T T T T F F T T T T F F T T F T T T T

For the middle statement to be true (the tilde is the main operator), the conditional must be false. That meanst A and B must be true and C false. This makes the third statement true, but the first statement false. It is impossible to make all of the statements true at the same time, so the statements are inconsistent.

4. A B (B C) & ~(A E) D & (E B) A E T T T T T T T F T T T T T T T T T T T

F T T T T T T T F F F T T F T T F T F

The biconditional will be true in two ways; if A and E are both true and if A and E are both false. If A and E are true, the second statement comes out false. But if A and E are both false, the statements all come out true. Since all we need is one line on which all of the statements are consistent, this line is enough to show that they’re consistent.

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Dr. Daniel R. Kern

5. A B (A C) & (B ~D) D C D & C T T T T T T T T T TF F T T F T T

F T F F T F T F T FT T T F F T F

The biconditional will be true if A and B are both true or if they are both false. If A and B are both true, then D & C comes out false. If A and B are both false, D C and D & C come out false. So there is not way to make all the statements true at the same time, so they are inconsistent.

The Indirect Method for testing for validity

1. (B & A) (A ~C) (A B) ~C (B & C) (A B) F F T T T F FT T T F T FT T T T F T F F

There is only one way for the conclusion to be false. On this truth-value assignment, the second premise is not true. So there are no truth value assignments on which the premises are true and the conclusion is false, so the argument is valid.

2. (A B) (B ~C) ((A & B) C) B (B A) ~C F F T T T F FT F F T T T T T T F F F FT

There is only one way for the conclusion to be false. On this truth value assignment, the premises all come out to be true. So there is a truth value assignment on which the premises are true and the conclusion is false, so the argument is invalid.

3. (A & C) B ~C D (D & B) C A D A B T F F T F TF T T T F F T F T T T T F F

F F T T T FT T F T T T T F T F F F T

The conclusion will be false if A is true and B false, or if A is false and B true. In the first case, one of the premises is also false. In the second case, however, all of the premises can be made true. So there is a case (one is all that is needed) in which the premises are true and the conclusion is false, so the argument is invalid.

4. A C B D E A (E & C) B D E A & B T T T F T T F T T F F T T F T T F T F F

F T T T F T F F F T T F F T

F T F T T F T F F F T F T T F F F F

The conclusion can be false in any of three ways. On each of these truth value assignments, a premise comes out false. Hence, there is no way to make the premises true while the conclusion is false, so the argument is valid. (NOTE: Any statements or statement letters that are not filled out will not change the result and so don’t matter).

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Dr. Daniel R. Kern

5. B C C (D & ~E) (D E) (A & G) G & H (H & D) C A B F T T T T T T TF T F F T T T T T T T T T T T T T F F

F T T

There are two ways for the conclusion to be false. On the first truth value assignment, all of the premises come out to be true. Since there is now a row on which the premises are all true and the conclusion is false, the argument can be declared invalid, and the second row need not be checked.

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Dr. Daniel R. Kern

Chapter 3

PART 1: Use truth trees to determine whether the following sets of statements are consistent or not. If they are, provide a set of truth values that will make them all true.

A. A & B (A B) C ~C

1. A & B Given2. (A B) C Given3. ~C Given4. A 1, &E5. B 1, &E

6. A B C 2, E

7. A B 6, E

The tree has at least one open branch, so it is an open tree, so the statements are consistent. A set of truth values that will make all of the statements true can be recovered from any open branch, so:

A B C T T F

B. A (B & C) C D D ~A

1. A (B & C) Given2. C D Given3. ~D A

4. ~C D 2, E

5. ~A B & C ~A B & C 1, E6. B B 5, &E7. C C 5, &E

8. ~D D ~D D ~D D 3, E9. A ~A A ~A A ~A 3, E


Dr. Daniel R. Kern

The tree has 3 open branches. Only one is needed to show that the set is consistent. Any of the three will provide a set of truth values on which all of the statements are true. From the left-most open branch:

A B C D F T/F F T

Note: Since B doesn’t show up on the branch, its truth value doesn’t matter on this assignment.

C. A & (B C) C ~(A D) ~(B ~E) ~(E ~C)

1. A & (B C) Given2. C ~(A D) Given3. ~(B ~E) Given4. ~(E ~C) Given5. A 1, &E6. (B C) 1, &E7. B 3, ~E8. E 3, ~E

9. B C 6, E

10. ~C ~(A D) ~C ~(A D) 2, E11. ~A ~A 10, ~E12 D D 10, ~E

13. E ~E 4, ~E14. C ~C 4, ~E

All of the branches on this truth tree close. This means that there are no truth value assignments on which all of the statements are true, which means that the set is inconsistent.


Dr. Daniel R. Kern

D. F ~(D & ~E) (F & H) & ~G D ~H H E

1. F ~(D & ~E) Given2. (F & H) & ~G Given3. D ~H Given4. H E Given5. F & H 2, &E6. F 5, &E7. H 5, &E8. ~G 2, &E

9. D ~H 3, E

10. ~H E 4, E

11. F ~F 1, E12. ~(D & ~E) D H 1, E

13. ~D E 12, ~&E

The tree has an open branch. This means that there is a truth value assignment under which all of the statements are true, which means the set is consistent. The truth value assignment under which the statements are true is:

D E F G H T T T F T


Dr. Daniel R. Kern

E.1. P ~(Q R) Given2. (~P Q) ~U Given3. ~(P & U) Given4. U ~W Given

5. ~(P Q) ~U 2, E6. ~P 5, ~E7. ~Q 5, ~E

8. P ~P P ~P 1, ≡E9. ~(Q R) Q R ~(Q R) Q R 1, ≡E

X

10. Q R Q R 9, EX

11. ~P ~U ~P ~U ~P ~U ~P ~U 3, ~&EX

12. U ~U 4, ≡E13. ~W W 4, ≡E

OThere is an open (completed) branch; the tree is open; the set is consistent

PART 2: Use truth trees to determine whether the following statements are logically true, logically false, or neither:

A. ~(W & E) (S N)

Step 1: Test the statement for logical falsehood:

1. ~(W & E) (S N) Given2.3. 4. ~(W & E) (S N) 1, E

5. ~W ~E 4, ~&E

Since the left branch is open, it is possible for the statement to be true, which would make the disjunction true (whatever the right disjunction is), so it is possible for the statement to be true, so it is NOT logically false.

Step 2: Test for logical truth – Test the negated statement.

1. ~[~(W & E) (S N)] Given2. ~~(W & E) 1, ~E3. ~(S N) 1, ~E4. W & E 2, ~~E5. W 4, &E6. E 4, &E7. ~S 3, ~E8. ~N 3, ~E

The negated statement has an open truth tree, so it is possible for it to be true, which means that it is possible for the non-negated statement to be false, which means that the statement is NOT a logical truth.The statement is neither logically false nor logically true, so it is contingent.

B. (A B) & (A & ~B)


1. (A B) & (A & ~B) Given2. A B 1, &E3. A & ~B 1, &E4. A 3, &E5. ~B 3, &E

6. A ~A 2, E7. B ~B 2, E


Dr. Daniel R. Kern

The statement has a closed truth tree. This means that it is not possible to make the statement true, which means that it is logically false.

C. (C D) ~(~C D)


1. (C D) ~(~C D) Given

2. ~(C D) ~(~C D) 1, E3. ~C 2, ~E4. ~D 1, ~E

The left branch of the tree is open, so the tree is open. The statement is not a contradiction.

Step 2: Test for logical truth – Test the negated statement.

1. ~[(C D) ~(~C D)] Given2. C D 1, ~E3. ~~(~C D) 1, ~E4. ~C D 3, ~~E

5. C D 2, E

6. ~~C D ~~C D 4, E7. C C 6, ~~E

The negated statement has at least one open branch (only one open branch is required). So the statement is not a tautology. The statement is neither logically true nor logically false, so it is contingent.


Dr. Daniel R. Kern

D. [A & (B C)] & [(~A ~B) & ~C]


1. [A & (B C)] & [(~A ~B) & ~C] Given2. A & (B C) 1, &E3. (~A ~B) & ~C 1, &E4. A 2, &E5. B C 2, &E6. ~C 3, &E7. ~A ~B 3, &E

8. B C 5, E

9. ~A ~B 7, E

The statement has a closed truth tree. This means that it is not possible to make the statement true, so it is logically false.

E. [B & (F C)] [(B & F) (B & C)]


1. [B & (F C)] [(B & F) (B & C)] Given

2. ~[B & (F C)] [(B & F) (B & C)] 1, E

3. ~B ~(F C) 2, ~&E

The left-most branch is open, so the tree is open, so the statement is not logically false.

NOTE: As soon as there is an open branch, there is no need to proceed further with the truth tree – ONE open branch is all that is needed for the proof.


Dr. Daniel R. Kern

Step 2: Test for logical truth – Test the negated statement

1. ~{[B & (F C)] [(B & F) (B & C)]} Given2. [B & (F C)] 1, ~E3. ~[(B & F) (B & C)] 1,~ E4. B 2, &E5. F C 2, &E6. ~(B &F) 3, ~E7. ~(B & C) 3, ~E

8. F C 5, E

9. ~B ~F ~B ~F 6, ~&E

10. ~B ~C 7, ~&E

The negated statement has a closed truth tree, which means that it cannot be made true. This means that the non-negated statement cannot be made false. So it is logically true.

Part 3: Use truth trees to determine whether the following statements are equivalent or not: (You don’t have to worry about this set)

A. A B ~A ~B

Test the negated biconditional: ~[(A B) (~A ~B)]

1. ~[(A B) (~A ~B)] Given

2. A B ~(A B) 1, ~E3. ~(~A ~B) ~A ~B 1, ~E4. ~A 3, ~E5. ~~B 3, ~E6. B 5, ~~E

7. ~A B 2, E

We have at least one open branch, so the tree is open. This means:1. It is possible to make the negated biconditional true;2. It is possible to make the conditional false;


Dr. Daniel R. Kern

3. There is at least one truth value assignment on which the two statement have different truth values;

4. The statements are NOT equivalent.

B. A B ~B ~A

Test the negated biconditional: ~[(A B) (~B ~A)]

1. ~[(A B) (~B ~A)] Given

2. A B ~(A B) 1, ~E3. ~(~B ~A) ~B ~A 1, ~E4. ~B 3, ~E5. ~~A 3, ~E6. A 5, ~~E

7. ~A B 2, E

8. A 2, ~E9. ~B 2, ~E

10. ~~B ~A 3, E11. B 10, ~~E

The truth tree for the negated biconditional is closed. This means:1. It is not possible to make the negated biconditional true;2. It is not possible to make the non-negated biconditional false;3. The two statements never have different truth values;4. The two statements are equivalent.


Dr. Daniel R. Kern

C. [B & (F C)] [(B & F) (B & C)]

Test the negated biconditional: ~{[B & (F C)] [(B & F) (B & C)]}

1. ~{[B & (F C)] [(B & F) (B & C)]} Given

2. [B & (F C)] ~[B & (F C)] 1, ~E3. ~[(B & F) (B & C)] [(B & F) (B & C) 1, ~E4. B 2, &E5. F C 2, &E6. ~(B & F) 3, ~E7. ~(B & C) 3, ~E

8. ~B ~F 6, ~&E

9. F C 5, E

10. ~B ~C 7, ~&E

11. ~B ~(F C) 2, ~&E12. ~F 11, ~E13. ~C 11, ~E

14. (B&F) (B&C) (B&F) (B&C) 3E15. B B B B 14, &E16. F C F C 14, &E

The truth tree for the negated biconditional is closed. This means that:

1. The negated biconditional cannot be made true;2. The non-negated biconditional cannot be made false;3. The two statements never have opposite truth values;4. The two statements are equivalent.


Dr. Daniel R. Kern

D. [P (Q R)] [(P & Q) R]

Test the negated biconditional: ~ [P (Q R)] [(P & Q) R]

1. ~[P (Q R)] [(P & Q) R] Given

2. [P (Q R)] ~[P (Q R)] 1, ~E3. ~[(P & Q) R] [(P & Q) R] 1, ~E4. P & Q 3, ~E

~R 3. ~E4. P 4, &E6. Q 4, &E

7. ~P Q R 2, E

8. ~Q R 7, E

9. P 2, ~E10. ~(Q R) 2, ~E11. Q 10, ~E12. ~R 10, ~E

13. ~(P & Q) R 3, E

14. ~P ~Q

The truth tree for the negated biconditional is closed. This means that:1. The negated biconditional cannot be made true;2. The non-negated biconditional cannot be made false;3. The two statements never have opposite truth values;4. The two statements are equivalent.


Dr. Daniel R. Kern

E. [A (B (C D))] [~A (B (~D ~C))]

Test the negated biconditional: ~{[A (B (C D))] [~A (B (~D ~C))]}

1. ~{[A (B (C D))] [~A (B (~D ~C))]} Given

2. [A (B (C D))] ~[A (B (C D))] 1, ~E3. ~[~A (B (~D ~C))] [~A (B (~D ~C))] 1, ~E4. ~~A 3, ~E5. ~(B (~D ~C)) 3, ~E6. A 4, ~~E7. ~B 5, ~E8. ~(~D ~C) 5, ~ E9. ~D 8, ~E10. ~~C 8 ~E11. C 10, ~~E

11. ~A (B (C D)) 2, E

12. ~B C D 11, E

The truth tree has an open branch. This means:1. It is possible to make the negated biconditional true;2. It is possible to make the non-negated biconditional false;3. There is at least one truth value assignment on which the two statement have different truth

values;4. The two statements are NOT equivalent.


Dr. Daniel R. Kern

Part 4: Use truth trees to test the following arguments for validity.

A. A ~B C ~D A C / ~B ~D

For the test, add the NEGATION of the conclusion to the set of premises and test the set for consistency.

1. A ~B Given2. C ~D Given3. A C Given4. ~(~B ~D) Given5. B 4, ~E6. D 4, ~E

5. A C 3, E

6. ~A ~B ~A ~B 1, E

7. ~C ~D 2, E

The truth tree is closed. This means:

1. The set of statements is inconsistent;2. It is not possible to make the premises true while the negation of the conclusion is true;3. It is not possible to make the premises true while the conclusion is false;4. The argument is valid.


Dr. Daniel R. Kern

B. A ~(B C) D C (D F) A / ~A B

1. A ~(B C) Given2. D C Given3. (D F) A Given4. ~(~A B) Given5. A 4, ~E6. ~B 4, ~E

7. ~A ~(B C) 1, E8. ~B 7, ~E9. ~C 7, ~E

10. ~D D 2, E11. ~C C 2, E

12. ~(D F) A 3, E

The truth tree is open. This means:

1. The set of statements is consistent;2. It is possible to make the premises true while the negation of the conclusion is true;3. It is possible to make the premises true while the conclusion is false;4. The argument is invalid.

NOTES:1. On lines 10 and 11 I reversed the normal order for the biconditional elimination. This is

because I saw that the one branch would close and I wanted to keep the tree branching toward the middle of the page.

2. I have not eliminated the negated disjunction on line 12. This is because the branch with the A on it is open and the branch with the negated disjunction won’t affect that branch, so the branch with the A on it is already a completed open branch, so there’s no need to proceed further.


Dr. Daniel R. Kern

C. D B (B & A) C ~C D / A D

1. D B Given2. (B & A) C Given3. ~C D Given4. ~(A D) Given5. A 4, ~E6. ~D 4, ~E

7. ~C D 3, E

8. D B 1, E

9. B&A ~(B &A) 2, E10. C ~C 2, E

11. ~B ~A 9, ~&E

The tree is closed. This means:

1. The set of statements is inconsistent;2. It is not possible to make the premises true while the negation of the conclusion is true;3. It is not possible to make the premises true while the conclusion is false;4. The argument is valid.


Dr. Daniel R. Kern

D. (A & B) C C ~(D E) (A D) (B ~C) / A & B

1. (A & B) C Given2. C ~(D E) Given3. (A D) (B ~C) Given4. ~(A & B) Given

5. ~A ~B 4, ~&E

6. A & B C A & B C 1, E7. A A 6, &E8. B B 6, &E

9. C ~C C ~C 2, E10. ~(D E) (D E) ~(D E) (D E) 2, E11. ~D ~D 10, ~E12. ~E ~E 10, ~E

13. ~(A D) (B ~C) ~(A D) (B ~C) 3, E14. A A 13, ~E15. ~D ~D




Dr. Daniel R. Kern

E. (F &I) J ~(H J) ~(I F) & H / (~F & G) & H

1. (F &I) J Given2. ~(H J) Given3. ~(I F) & H Given4. ~[(~F & G) & H] Given5. ~(I F) 3, &E6. H 3, &E7. I 5, ~E8. ~F 5, ~E

9. ~(F & I) J 1, E

10. ~I ~F 7, ~&E

11. H ~H H ~H 2, ~E~J J ~J J 2, ~E

12. ~(~F & G) ~H 4, ~&E

13. F ~G 12, ~E



Chapter 4

Regular Proofs

A. Derive: F (5 steps)1. (A & B) & C Given2. B D Given3. (D G) F Given 4. (A & B) 1, &E5. B 4, &E6. D 2, 5, MP7. D G 7, I8. F 3, 8 MP

B. Derive: P (4 steps)1. A Given2. (A Z) (B & ~G) Given3. ~G P Given 4. A Z 1, I5. B & ~G 5, 7, BMP6. ~G 8, &E7. P 3, 8, MP

C. Derive: ~D (5 steps)1. A Given2. [A (X & (Y Z))] B Given3. B ~(C & D) Given4. C Given 5. A (X & (Y Z)) 1, I6. B 2, 5, MP7. ~(C & D) 3, 6, MP8. ~C ~D 8, DeM9. ~D 4, 8, DS

D. Derive: ~L (5 steps)1. (L K) (A & B) Given2. ~A Given3. K A Given 4. ~A ~B 2, I5. ~(A & B) 4, DeM6. L K 1, 5, DS7. ~K 2, 3, MT8. ~L 6, 7, MT

E. Derive J (8 steps)1. (~A ~B) & (A E) Given2. B Given3. (E H) (~B J) Given 4. ~A ~B 1, &E5. B A 5, Trans.6. A 2, 5, MP7. A E 1, &E8. E 6,7 , MP9. E H 8, I.10. ~B J 3, 9, MP11. J 2, 10, DS

F. Derive: ~O (5 steps)1. (M N) Given2. (M & ~N) (O P) Given3. ~P Given 4. (M N) 1, BMP5. (~M N) 4, MI6. ~(M & ~N) 5, DeM7. O P 2, 6, DS8. ~O 3, 8, MT

G. Derive: G (5 steps)1. (D & E) (F & G) Given2. F C Given3. ~C Given4. (D & E) G Given 5. ~F 2, 3, MT6. ~F ~G 5, I7. ~(F & G) 6, DeM8. D & E 1, 7, DS9. G 8, 4, MP


Dr. Daniel R. Kern

H. Derive: (I J) (8 steps)1. (~F ~G) (I J) Given2. B F Given3. ~G D Given4. C & (B & ~D) Given 5. B & ~D 4, &E6. B 5, &E7. ~D 5, &E8. F 2, 6, MP9. G 3, 7, MT10. F & G 8, 9 &I11. ~(~F ~G) 10, DeM12. I J 1,11, DS

I. Derive: B (8 or 9 steps)1. A ~(B & ~C) Given 2. ~A D Given3. (B C) (~F B) Given4. F & ~D Given 5. ~D 4, &E6. ~~A 2, 5 MT7. A 6, DN8. ~(B & ~C) 1, 7, MP9. ~B C 8, DeM10. B C 9, MI11. ~F B 3, 10, MP12. F 4, &E13. B 11,12, DS

J. Derive: B (8 steps)1. (C D) Given2. ~(C & D) Given3. (C D) (A F) Given4. ~(F C) Given5. (A B) Given 6. (C & D) (~C & ~D) 1, ME7. (~C & ~D) 2, 6, DS8. ~(C D) 7, DeM9. A F 3, 8, DS10. ~F & ~C 4, DeM11. ~F 10, Simp12. A 9,11, DS13. B 12, 14, MP


Dr. Daniel R. Kern

Conditional Proofs

A. Derive: A B1. A (D H) Given2. ~A ~D Given3. H (B ~A) Given4. A Assume5. D H 1, 4, MP6. ~D 2, 4, DS7. H 5, 6, DS8. B ~A 3, 7, BMP9. B 4, 8, DS10. A B 4- 9, CP

B. Derive (A B) D1. ~A & C Given2. (C & B) ~H Given3. (F N) H Given4. ~F D Given5. A B Assume6. ~A 1, &E7. B 5, 6, DS8. C 1, &E9. C & B 7, 8, &I10. ~H 2, 9, BMP11. ~(F N) 3,10, MT12. ~F & ~N 11, DeM13. ~F 12, &E14. D 4,13, BMP15. (A B) D

C. Derive: A (B C)1. (A & B) D Given2. D ~G Given3. G C Given4. A Assume5. B Assume6. A & B 4, 5, &I7. D 1, 6, MP8. ~G 2, 7, BMP9. C 3, 8, DS10. B C 5-9, CP11. A (B C) 4-10, CP

D. Derive: A (B (C D))1. (A & B) F Given2. ~(C & G) Given3. ~G (D ~F) Given4. A Assume5. B Assume6. C Assume7. A & B 4, 5, &I8. F 1, 7, MP9. ~C ~G 2, DeM10. ~G 6, 9, DS11. D ~F 3, 10, BMP12. D 8, 11, DS13. C D 6-12, CP14. B (C D) 5-13, CP15. A (B (C D)) 4-14, CP

E. Derive: A B1. (A C) D2. ~F ~D3. ~(F & G)4. G B5. ~(~B & G) F6. ~F A7. A Assume8. A C 7, I9. D 1, 8, BMP10. F 2, 9, MT11. ~F ~G 3, DeM12. ~G 10, 11, DS13. B 4, 12, DS14. A B 7-13, CP15. B Assume16. (B ~G) F 5, DeM17. B ~G 15, I18. F 16, 17, MT19. A 6,19, DS20. B A 15-19, CP21. (A B) & (B A) 14, 20, &I22. A B 12, ME


Dr. Daniel R. Kern

Indirect Proofs

A. Derive: A BNOTE: There are two ways to solve this – one derives a contradiction with B, one derives a contradiction with D. Both are valid proofs; they are both demonstrated here.

1. ~A C2. (~B F) D3. (C & D) A4. ~(A B) Assume 5. ~A & ~B 4, DeM6. ~A 5, &E7. C 1, 6, MP8. ~(C & D) 3, 6, MT9. ~C ~D 8, DeM10. ~D 7, 9, DS11. ~(~B F) 2, 10, MT12. B & ~F 11, DeM13. B 12, &E14. ~B 5, &E15. B & ~B 13, 4, &I16. A B 4-15, IP

1. ~A C Given2. (~B F) D Given3. (C & D) A Given4. ~(A B) Assume5. ~A & ~B 4, DeM6. ~B 5, &E7. ~B F 6, I8. D 2,7, BMP9. ~A 5, &E10. C 1,9, MP11. ~(C & D) 3,9, MT12. ~C ~D 11, DeM13. ~D 10,12, DS14. D & ~D 8,14, &I15. A B 4-14, IP

B. Derive: ~(B G)NOTE: On this proof, lines 4-8 can all be derived in the scope of the main proof, without reference to the assumption. It would also be possible to assume B G on line 4 and include all of these lines within the scope of the indirect proof. Either way is fine here.

1. B & ~H Given2. H (F I) Given3. (~F & ~I) ~G Given4. B 1, &E5. ~H 1, &E6. F I 2, 5, DS7. ~(F I) ~G 3, DeM8. ~G 6, 7, DS9. B G Assume10. G 4, 9, BMP11. G & ~G 8, 10, &I12. ~(B G) 9-11, IP

C. Derive: ~[M & (N O)]1. N ~M2. (O R) ~(H G)3. M H4. M & (N O) Assume5. M 4, &E6. H 3, 5, MP7. ~N 1, 5, MP8. N O 4, &E9. O 7,8, DS10. O R 9, I11. ~(H G) 2, 10, BMP12. ~H & ~G 11, DeM13. ~H 12, &E14. H & ~H 6,13, &I15. ~[M & (N O)] 4-14, IP


Dr. Daniel R. Kern

D. Derive: A B1. A ~(C D) Given2. F D Given3. (C & ~F) (G H) Given4. H F Given5. ~C ~G Given6. ~(A B) Assume7. ~A & ~B 6, DeM8. ~A 7, &E9. ~(C D) 1, 8, DS10. ~(~C D) 9, MI11. C & ~D 10, DeM12. ~D 11, &E13. ~F 2, 12, BMT14. C 11, &E15. C & ~F 13,14, &I16. G H 3, 15, MP17. ~H 4, 13, MT18. G 16,17, DS19. ~G 5, 14, DS20. G & ~G 18,19, &I21. A B 6-20, IP

E. Derive (B C) (B & ~C)1. ~(B C) (I J) Given2. (B C) ~J Given3. ~I (~J M) Given4. I N Given5. (M & N) I Given6. ~[(B C) (B & ~C)] Assume7. ~(B C) & ~(B & ~C) 6, DeM8. ~(B C) 7, &E9. ~(B & ~C) 7, &E10. I J 1, 8,

BMP11. (~B C) ~J 2, MI12. (B & ~C) ~J 11,

DeM13. ~J 9, 12,

DS14. ~I 10,

13, MT15. N 4, 14,

DS16. ~(M & N) 5, 14,

MT17. ~M ~N 16,

DeM18. ~M 15, 17,

DS19. ~J M 3, 14,

MP20. J 18, 19,

BMT21. J & ~J 13, 20,

&I22. (B C) (B ~C) 6-21, IP


Dr. Daniel R. Kern

Chapter 5

Using the following predicate symbols and individuals, translate the sentences:

Lxy = x likes yFxy = x is from yDx = x dances Wxy = x sings with ySx = x singsa = Albertb = Bobc = Carmeliad = Doral = Los Angelesm = Memphis

Translate:

1. Albert sings and Albert dances, but Albert is not from Memphis.(Sa & Da) & ~Fam

2. If Albert sings with Carmelia, then Albert does not sing with Bob and Bob does not sing with Carmelia.Wxy (~Wab & ~Wbc)

3. If Bob is from Los Angeles, then Caremelia likes him, but if Bob is from Memphis, then Carmelia doesn’t like him.(Fbl Lcb) & (~Fbm ~Lcb)

4. Dora is from Memphis and she sings and dances, but neither Albert nor Bob like her.(Fdm & (Sd & Dd)) & (~Lad & ~Lbd)

5. Albert sings with Dora and Bob sings with Carmelia, but Albert doesn’t sing with both Bob and Carmelia. (Wad & Wbc) & (~Wab ~Wac)

6. Albert, Dora, and Bob are all from Los Angeles, but Carmelia is from Memphis.(Fal & (Fdl & Fbl)) & Fcm

7. Either Albert is from Los Angeles and sings and dances, or Dora is from Memphis and doesn’t sing or doesn’t dance.(Fal & (Sa & Da)) (Fdm & (~Sd & ~Dd))

8. One of the four people is from Los Angeles.(Fal Fbl) (Fcl Fdl)


Dr. Daniel R. Kern

9. All of the four people are from Memphis.(Fam & Fbm) & (Fcm & Fdm)

10. The four people are not all from Los Angeles, and none of the four people is from Memphis.~[(Fal Fbl) (Fcl Fdl)] & ~[(Fam & Fbm) & (Fcm & Fdm)]


Dr. Daniel R. Kern

Chapter 6

Single-place predicatesUsing the specified universe of discourse and predicates, translate the following statements into quantificational form. In the cases where there are more than one possible symbolization, this is indicated, with the most direct symbolization given first.

UD: Humans.

Mx = X has a motherSx = X has a sisterBx = X has a brotherCx = X has a child

1. (x)Mx

2. ~(x)Cx OR (x)~Cx

3. (x)Sx & (x)Bx

4. (x)Cx & (x)Mx

5. (x)~Cx & ~(x)~Mx

6. (x)Cx (x)Sx

7. (x)Cx (y)My

8. (x)(Bx & Sx)

9. ~(x)(Bx v Sx) OR (x)~(Bx v Sx) OR (x)(~Bx & ~Sx)

10. ~(x)(Bx & Sx) OR (x)~(Bx & Sx) OR (x)(~Bx v ~Sx)

11. (x)~(Bx v Sx) OR (x)(~Bx & ~Sx) OR ~(x)(Bx v Sx)

12. (x)(Mx & ~(Bx v Sx)) OR (x)(Mx & (~Bx & ~Sx))

13. (x)(Mx & (Bx Sx))

14. (x)[(Cx Mx) & (Sx Bx)]

Multi-place predicatesUsing the specified universe of discourse and predicates, translate the following statements into quantificational form (a number of these statements are false; don’t worry about that at this point):

UD: Humans

Lxy = x likes yMxy = x misses y


Dr. Daniel R. Kern

Ox = x is out of town

1. (x)(y)Lxy

2. (x)(y)Lxy

3. ~(x)(y)Lxy OR (x)(y)~Lxy

4. ~(x)(y)~Lxy OR (x)(y)Lxy

5. (x)(y)Lxy OR (x)~(y)~Lxy

6. (x)(y)(Lxy & Mxy)

7. (x)(y)(Mxy Lxy)

8. (x)(y)(~Mxy ~Lxy)

9. (x)(y)((Lxy & Oy) Mxy)

10. (x)(y)((~Lxy & Oy) ~Mxy)

Multi-place predicates with expanded UDUsing the specified predicates, and no specified universe of discourse, translate the following statements into quantificational form.

Lxy = x likes ySxy = x sits on y’s lapFxy = x licks y’s facePx = x is a personDx = x is a dogTx = x wags its tailMx = x is a maleFx = x is a female

1. (x)(y)((Px & Py) ~Lxy)

2. ~(x)(y)((Px & Py) ~Lxy) OR

(x)(y)((Px & Py) & Lxy)

3. ~(x)(y)((Px & Py) & Lxy) OR

(x)(y)((Px & Py) ~Lxy)

4. (x)(y)[((Px & Mx) & (Py & Fy)) Lxy]

5. (x)(y)[((Px & Mx) & (Py & Fy)) & Lxy]

6. (x)(y)[(Px & Mx) & (Py & Fy) ~Lxy] OR

(x)(y)[(Px & Mx) > ((Py & Fy) ~Lxy)] (these are the same by Exportation)

7. (x)(y)[((Px & Fx) & Lxy) (Py & Fy)] OR

8. (x)(y)[((Px & Fx) & ~(Py & Fy)) ~Lxy]

9. (x)(y)[Px & (Dy Lxy)]


Dr. Daniel R. Kern

10. (x)(y)[Dx & (Py ~Lxy)]

11. (x)(y)[Px & (Dy ~Lxy)]

12. (x)(y)[(Px & ((Dy v Py) ~Lxy)]

13. (x)(y)[((Dx & Py) & Lxy) (Tx & (Fxy & Sxy))]

14. (x)(y)[((Dx & Mx) & (Dy & Fy)) Fxy]

Chapter 7

Translate the following sets of statements and use truth trees to test them for consistency.

UD: EverythingAx: x is an animalBx: x is baldCx: x is a politicianDx: x is a DemocratHx: x is honestLx: x has lungsMx: x is from MississippiPx: x is a personWx: x has wingsTx: x is from Texasf: Francois

A. Some people are bald. Some people are not bald.

1. (x)(Px & Bx) Given2. (x)(Px & ~Bx) Given3. Pa & Ba 1, I4. Pb & ~Bb 2, I5. Pa 3, &E6. Ba 3, &E7. Pb 4, &E8. ~Bb 4, &E

The tree is open. The set is consistent.

B. Some animals have lungs. Some animals do not have lungs.

1. (x)(Ax & Lx) Given2. (x)(Ax & ~Lx) Given3. Aa & La 1, I4. Ab & ~Lb 2, I5. Aa 3, &E6. La 3, &E7. Ab 4, &E8. ~Lb 4, &E

The tree is open. The set is consistent.


Dr. Daniel R. Kern

C. Some animals have lungs. Some animals have wings. Some animals are bald.

1. (x)(Ax & Lx) Given2. (x)(Ax & Wx) Given3. (x)(Ax & Bx) Given4. Aa 1, I5. La 1, I6. Ab 2, I7. Wb 2, I8. Ac 3, I9. Bc 3, I

D. All people have lungs. Some people do not have lungs.

1. (x)(Px Lx) Given2. (x)(Px & ~Lx) Given3. Pa & ~La 1, I4. Pa La 2, I5. Pa 3, &E6. ~La 3, &E

7. ~Pa La 4, E

Both branches close. The set is inconsistent.

E. All animals have lungs. Some animals have wings. Some animals don’t have wings.

1. (x)(Ax Lx) Given2. (x)(Ax & Wx) Given3. (x)(Ax & ~Wx) Given4. Aa & Wa 2, I5. Ab & ~Wb 3, I6. Aa 4, &E7. Wa 4, &E8. Ab 5, &E9. ~Wb 5, &E

10. ~Aa La 1, I

11. ~Ab Lb 1, I


Dr. Daniel R. Kern

The tree is open; the set is consistent. Note that we instantiated the universal for EACH variable that was introduced on the branches above it, to be sure we had covered all the possibilities.

F. It is not the case either that all animals have lungs or that all animals have wings. Some animals have wings and lungs. Some animals have neither wings nor lungs.

1. ~[(x)(Ax Lx) (x)(Ax Wx)] Given2. (x)(Ax & (Wx & Lx)) Given3. (x)(Ax & (~Wx & ~Lx)) Given4. ~(x)(Ax Lx) 1, ~E5. ~(x)(Ax Wx) 1, ~E6. (x)~(Ax Lx) 4 ~E7. (x)~(Ax Wx) 5, ~E 8. Aa & (Wa & La) 2, I9. Ab & (~Wb & ~Lb) 3, I10. Wa & La 8, &E11. ~Wb & ~Lb 9, &E12. Aa 8, &E13. Ab 9, &E14. Wa 10, &E15. La 10, &E16. ~Wb 11, &E17. ~Lb 11, &E18. ~(Ac Lc) 6, I19. Ac 18, ~E20. ~Lc 18, ~E21. ~(Ad Wd) 7, I 22. Ad 21, ~E23. ~Wd 21, ~E

The tree is open; the set is consistent. The order of operations here can be done in a number of different ways; the result is the same.


Dr. Daniel R. Kern

G. All politicians are dishonest. All Mississippians are honest. Some politicians are from Mississippi.

1. (x)(Cx ~Hx) Given2. (x)(Mx Hx) Given3. (x)(Cx & Mx) Given4. Ca & Ma 3, I5. Ca ~Ha 1, I6. Ma Ha 2, I7. Ca 4, &E8. Ma 4, &E

9. ~Ca ~Ha 5, E

10. ~Ma Ha 6, E

The tree is closed; the set is inconsistent.

H. Some Democratic politicians are Mississippians. Some Democratic politicians are dishonest. No Mississippians are dishonest.

1. (x)(Dx & (Cx & Mx)) Given2. (x)(Dx & (Cx & ~Hx)) Given3. ~(x)(Mx & ~Hx) Given OR4. (x)(Mx Hx) Given (not used)5. (x)~(Mx & ~Hx) 3, ~E6. Da & (Ca & Ma) 1, I7. Db & (Cb & ~Hb) 2, I8. Da 6, &E9. Ca & Ma 6, &E10. Ca 9, &E11. Ma 9, &E12. Db 7, &E13. Cb & ~Hb 7, &E14. Cb 13, &E15. ~Hb 13, &E16. ~(Ma & ~Ha) 5, x 17. ~(Mb & ~Hb) 5, x

18. ~Ma Ha 16, ~&E

19. ~Mb Hb 17, ~&E

The tree is open; the set is consistent.


Dr. Daniel R. Kern

I. Francois is either a Texan or a Mississippian. Francois is a politician. Francois is bald. All Texans are dishonest. All politicians are dishonest. There are no dishonest bald people.

1. Tf Mf Given2. Cf Given3. Bf Given4. (x)(Tx ~Hx) Given5. (x)(Cx ~Hx) Given6. ~(x)(Bx & ~Hx) Given7. (x)~(Bx & ~Hx) 6, ~I

8. Tf Mf 1, E9. Tf ~Hf Tf ~Hf 4, E

10. ~Tf ~Hf ~Tf ~Hf 9, E

11. Cf ~Hf Cf ~Hf Cf ~Hf 5, I

12. ~Cf ~Hf ~Cf ~Hf ~Cf ~Hf 11, E

13. ~(Bf & ~Hf) ~(Bf & ~Hf) ~(Bf & ~Hf) 7, I

14. ~Bf Hf ~Bf Hf ~Bf Hf 13, ~&E

All branches close. The set is inconsistent.

J. If some Texans are not politicians, then all Texans are honest. If some dishonest people are from Mississippi, then no politicians are from Mississippi. Some dishonest people are from Mississippi.

1. (x)(Tx & ~Cx) (x)(Tx ~Hx) Given2. (x)(~Hx & Mx) ~(x)(Cx & Mx) Given3. (x)(~Hx & Mx) Given4. ~Ha & Ma 3, I5. ~Ha 4, &E6. Ma 4, &E

7. ~(x)(Tx & ~Cx) (x)(Tx Hx) 1, E8. (x)~(Tx & ~Cx) 7, ~E9. ~(Ta & ~Ca) 8, I

10. ~Ta Ca

11. ~(x)(~Hx & Mx) ~(x)(Cx & Mx) ~(x)(~Hx & Mx) ~(x)(Cx & Mx) ~(x)(~Hx & Mx) ~(x)(Cx & Mx) 2, I12. (x)~(~Hx & Mx) (x)~(~Hx & Mx) (x)~(~Hx & Mx) 12, ~E13. ~(~Ha & Ma) ~(~Ha & Ma) ~(~Ha & Ma) 13, I

14. Ha ~Ma Ha ~Ma Ha ~Ma 14, ~&E

15. (x)~(Cx & Mx) (x)~(Cx & Mx) (x)~(Cx & Mx) 12, ~E16. ~(Ca & Ma) ~(Ca & Ma) ~(Ca & Ma) 16, I

17. ~Ca ~Ma 17, ~&E

The tree is open; the set is consistent. We haven’t completed the branches on the right side of the tree, but we don’t need to. A single completed open branch is all we need to show consistency.

Part 2: Translate the following arguments and use truth trees to test them for validity (the question of soundness (the truth of the premises) is not at issue here; only the question of validity).

UD: EverythingAx: x is an animalBx: x is baldLx: x has lungsPx: x is a personWx: x has wingsf: Francois

A. All people have lungs. Francois is a person. Therefore, Francois has lungs.

1. (x)(Px Lx) Given2. Pf Given3. ~Lf Assumed4. Pf Lf 1, I

5. ~Pf Lf 4, E

The tree is closed. The argument is valid.

B. All people have lungs. Francois has lungs. Therefore, Francois is a person.

1. (x)(Px Lx) Given2. Lf Given3. ~Pf Assumed4. Pf Lf 1, I

5. ~Pf Lf 4, E

The tree is open. The argument is invalid.


Dr. Daniel R. Kern

C. Some animals have wings. Some animals have lungs. Therefore, some animals have both wings and lungs.

1. (x)(Ax & Wx) Given2. (x)(Ax & Lx) Given3. ~(x)(Ax & (Wx & Lx)) Assumed4. (x)~(Ax & (Wx & Lx)) 3, ~E5. Aa & Wa 1, I6. Ab & Lb 2, I7. ~(Aa & (Wa & La)) 4, I8. ~(Ab & (Wb & Lb)) 4, I9. Aa 5, &E10. Wa 5, &E11. Ab 6, &E12. Lb 6, &E

13. ~Aa ~(Wa & La) 7, ~&E

14. ~La ~Wa 13, ~&E

15. ~Ab ~(Wb & Lb) 8, ~&E

16. ~Wb ~Lb 15, ~&E



Dr. Daniel R. Kern

D. All animals have lungs. All animals have wings. Therefore, all animals have wings and lungs.

1. (x)(Ax Wx) Given2. (Ax)(Ax Lx) Given3. ~(x)(Ax (Wx & Lx)) Assumed4. (x)~(Ax (Wx & Lx)) 3, ~E5. ~(Aa (Wa & La)) 4, I6. Aa Wa 1, I7. Aa La 2, I8. Aa 5, ~E9. ~(Wa & La) 5, ~E

10. ~Aa Wa 6, E

11. ~Aa La 7, E

12. ~Wa ~La 9, ~&E

The tree is closed. The argument is valid. This might seem strange, but remember, the issue here is the form of the argument, not the truth of anything. Both premises are false, so the argument is unsound, but formally, it is valid.


Dr. Daniel R. Kern

UD: PeopleCx: x is a politicianDx: x is a DemocratHx: x is honestMx: x is from MississippiTx: x is from Texasf: Francois

E. All politicians are dishonest. Some politicians are from Mississippi. Therefore, some Mississippians are dishonest.

1. (x)(Cx ~Hx) Given2. (x)(Cx & Mx) Given3. ~(x)(Mx & ~Hx) Assumed4. (x)~(Mx & ~Hx) 3, ~E5. Ca & Ma 2, I6. Ca 5, &E7. Ma 5, &E8. Ca ~Ha 1, I9. ~(Ma & ~Ha) 4, I

10. ~Ca ~Ha 8, E

11. ~Ma Ha 10, ~&E

The tree is closed. The argument is valid.


Dr. Daniel R. Kern

F. Some politicians are dishonest. Some Mississippians are politicians. Therefore, some Mississippians are dishonest.

1. (x)(Cx & ~Hx) Given2. (x)(Mx & Cx) Given3. ~(x)(Mx & ~Hx) Assumed4. (x)~(Mx & ~Hx) 3, ~E5. Ca & ~Ha 1, I6. Mb & Cb 2, I7. ~(Ma & ~Ha) 4, I8. ~(Mb & ~Hb) 4, I9. Ca 5, &E10. ~Ha 5, &E11. Mb 6, &E12. Cb 6, &E

13. ~Ma Ha 7, ~&E

14. ~Mb Hb 8, ~&E



Dr. Daniel R. Kern

G. No one from Texas or Mississippi is honest. Some dishonest politicians are from Mississippi. Therefore, no Texans are politicians.

1. (x)((Tx Mx) ~Hx) Given2. (x)((Cx & ~Hx) & Mx) Given3. ~(x)(Tx ~Cx) Assumed4. (x)~(Tx ~Cx) 3, ~E5. (Ca & ~Ha) & Ma 2, I6. ~(Tb ~Cb) 4, I7. Ca 5, &E8. ~Ha 5, &E9. Ma 5, &E10. (Ta Ma) ~Ha 1, I11. (Tb Mb) ~Hb 1, I12. Tb 6, ~E13. Cb 6, ~E

14. ~(Ta Ma) ~Ha 10, E15. ~Ta 14, ~E16. ~Ma 14, ~E

17. ~(Tb Mb) ~Hb 11, E18. ~Tb19. ~Mb


H. Some Texan politicians are honest. No honest politicians are from Mississippi. Some honest people are from neither Texas nor Mississippi. Therefore, some politicians are not from either Texas or Mississippi.1. (x)((Cx & Tx) & Hx) Given2. (x)((Cx & Hx) ~Mx) Given3. (x)(Hx & (~Tx & ~Mx)) Given4. ~(x)(Cx & (~Tx & ~Mx)) Assumed5. (x)~(Cx & (~Tx & ~Mx)) 4, ~E6. Ca & Ta & Ha 1, I7. Hb & (~Tb & ~Mb) 3, I8. Ca 6, &E9. Ha 6, &E10. ~Tb 7, &E11. ~Mb 7, &E12. (Ca & Ha) ~Ma 2, I13. (Cb & Hb) ~Mb 2, I14. ~(Ca & (~Ta & ~Ma)) 5,I15. ~(Cb & (~Tb & ~Mb)) 5,I

16. ~Ca ~(~Ta & ~Ma) 16, ~&E

17. Ta Ma 18, ~&E

18. ~Cb ~(~Tb & ~Mb) ~Cb ~(~Tb & ~Mb) 17, ~&E

19. Tb Mb Tb Mb 20, ~&E

20. ~(Ca & Ha) ~Ma ~(Ca & Ha) ~Ma 14, E

21. ~Ca ~Ha ~Ca ~Ha 22, ~&Ea

22. ~(Cb & Hb) ~Mb 15, E

The tree is open. The argument is invalid


Dr. Daniel R. Kern

I. Francois is either a Texan or a Mississippian. Francois is a politician. Francois is bald. All Texans are dishonest. All politicians are dishonest. Therefore, there are some dishonest bald people.

1. Tf Mf Given2. Cf Given3. Bf Given4. (x)(Tx ~Hx) Given5. (x)(Cx ~Hx) Given6. ~(x)(Bx & ~Hx) Assumed7. (x)~(Bx & ~Hx) 6, ~E8. Tf ~Hf 4, I9. Cf ~Hf 5, I10. ~(Bf & ~Hf) 7, I

11. ~Cf ~Hf 9, E

12. ~Bf Hf 10, ~&E

The tree is closed. The argument is valid. Note on this one that several statements have not been decomposed. This is fine, because there are no more open branches to decompose them on. This shows that there was more information in the premises than was needed to derive the conclusion.


Dr. Daniel R. Kern

J. If some Texans are not politicians, then all Texans are honest. If some dishonest people are from Mississippi, then no politicians are from Mississippi. Therefore, some dishonest people are not from Mississippi.

1. (x)(Tx & ~Cx) (x)(Tx Hx) Given2. (x)(~Hx & Mx) (x)(Cx ~Mx) Given3. ~(x)(~Hx & ~Mx) Assumed4. (x)~(~Hx & ~Mx) 3, ~E

5. ~(x)(Tx & ~Cx) (x)(Tx Hx) 1, E6. (x)~(Tx & ~Cx) 5, ~E

7. ~(x)(~Hx & Mx) (x)(Cx ~Mx) ~(x)(~Hx & Mx) (x)(Cx ~Mx) 2, E8. (x)~(~Hx & Mx) (x)~(~Hx & Mx) 7, ~E

9. ~(Ta & ~Ca) ~(Tx & ~Ca) 6, I10. ~(~Ha & Ma) 8, I

11. ~Ta Ca 9, ~&E

12. Ha ~Ma 10, ~&E

The tree is open. The argument is invalid. Although relatively little of the tree has been completed, the branches on the left are completed open branches – there are no more sentences to decompose on the branches above them. A single open branch is all that is needed to show consistency, which, in this case, shows invalidity.

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